We are given that the concentration of NaOH is 0.0003 M and are asked to calculate the pH
We know that NaOH dissociates by the following reaction:
NaOH → Na⁺ + OH⁻
Which means that one mole of NaOH produces one mole of OH⁻ ion, which is what we care about since the pH is affected only by the concentration of H⁺ and OH⁻ ions
Now that we know that one mole of NaOH produces one mole of OH⁻, 0.0003M NaOH will produce 0.0003M OH⁻
Concentration of OH⁻ (also written as [OH⁻]) = 3 * 10⁻⁴
<u>pOH of the solution:</u>
pOH = -log[OH⁻] = -log(3 * 10⁻⁴)
pOH = -0.477 + 4
pOH = 3.523
<u>pH of the solution:</u>
We know that the sum of pH and pOH of a solution is 14
pH + pOH = 14
pH + 3.523 = 14 [subtracting 3.523 from both sides]
pH = 10.477
Sand only bc the salt dissolved and sand doesn't
Answer:
Maybe or maybe not (not sure)
Explanation:
A displacement reaction is a type of reaction where one element is displaced by another from a compound.
In the case of magnesium and lead nitrate, magnesium is more reactive than lead. Therefore, it will displace lead from lead nitrate to form magnesium nitrate and lead.
The reaction can be represented as:
Mg(s) + Pb(NO3)2(aq) → Mg(NO3)2(aq) + Pb(s)
Another answer could be;
A displacement reaction does not take place in 'magnesium + lead nitrate' because magnesium is more reactive than lead.
Answer:
uy pasagot kung nasagutan mo na pls wala pa akong sagot jan
Explanation:
Since, it is given that critical temperature of Argon is 150.9 K and critical pressure of Argon is 48.0 atm.
It is known that gas phase of neon occurs at 50 K. As the boiling point of Ar is more than the boiling point of neon which means that there is strong intermolecular force of attraction between argon molecules as compared to neon molecules.
This is also because argon is larger in size. As a result, induced dipole-induced dipole forces leads to more strength in Ar as compared to Ne.
